FIN HEAT TRANSFER CLASS NOTES FOR MECHANICAL ENGINEERING - ME Subjects

FIN HEAT TRANSFER CLASS NOTES FOR

MECHANICAL ENGINEERING

Fin is a solid extended surface or a combined conduction convection system. Heat is transferred by Conduction within the solid and by convection from the fin surface in a PERPENDICULAR direction to that of conduction. Finally heat is lost to the surroundings. Fins exponentially increase the rate of heat transfer by increasing the surface area without increase of primary surface area. This increased surface area reduces the convective resistance and increases the rate of heat transfer economically. The fins should have large surface area per unit volume (>700 m²/m³) with small gas flow passages with natural laminar flow for increased rate of heat transfer. It increases the heat exchange between two fluids which are separated by a sold surface as in home radiators and cooling system in cars.

One-dimensional steady state heat conduction
Homogeneous and isotropic fin material, k is constant
No internal heat generation
Uniform cross-sectional area
Uniform convective heat transfer coefficient h
Negligible contact thermal resistance
Negligible radiation
Always calculate Rate of heat transfer at the base of the fin

Base of the fin or root of the fin: Where fin connects to the main body
Tip of the fin—The other free end of the fin
A_c is cross sectional area of the fin = b x t for a rectangular fin
A_sis the surface area of the fin= perimeter x length= (2b+2t) L

DERIVATION

FIN HEAT TRANSFER CLASS NOTES FOR

MECHANICAL ENGINEERING

PURPOSE OF A FIN: To increase heat transfer by increasing the surface area without increase of primary surface area. FIN involves both (CONDUCTIVE + CONVECTIVE) HEAT TRANSFER. . Finally it loses heat to the atmosphere.

HOW FINS INCREASE HEAT TRANSFER: Convection equation

q =h A dT Watts.

Undoubtedly heat transfer can be increased by increasing the

(i) temperature difference

(ii) the convection heat transfer coefficient

(iii) the contact surface area of the object

TYPE OF FINS / SHAPES OF FINS

(i)Fins of constant area of cross section like rectangular, circular, square and elliptical fins

(ii) Fins with varying area of cross section like triangular, trapezoidal , parabolic and conical

(iii) Plate fins and wound fins from the two types above

Fig. Types of fins

PURPOSE OF DESIGNING A FIN

Fin should be most effective. It should be able to transfer many times the heat as compared without fins. It should also be cost effective.

Write the assumption used in the analysis of heat transfer through fins.

FIN PARAMETER

Fin parameter is represented by the letter ‘m’. It is given by the relation given below

m = (h P/k Ac)0.5

Where

h surrounding medium convective HT coefficient

P is the perimeter of the fin

k thermal conductivity of fin material

Ac cross sectional area of the fin if rectangular, Ac =b x t

Q. FIN EFFICIENCY FOR A SINGLE FIN

It is applicable for a single fin.

Ratio of the actual rate of heat transfer to the maximum rate of heat transfer through the fin. Maximum heat transfer occurs if the entire fin were at the base temperature.

The fin efficiency is ηf = actual q.fin/ q.max with base temperature all along the length

ηf = (PhAck)1/2(Tb–T )/hPL (Tb–T )

= (kAc/hP)1/2(1/L)

=1/mL

Finally ηf =1/mL

Fin efficiency will always be less than one. This is

because assuming the temperature throughout

the fin is at the base temperature would increase

the heat transfer rate.

PRACTICAL APPLICATIONS OF FINS

(i) Electric motors

(ii) Car radiators

(iii) Pumps

(iv) Compressors

(v) Internal Combustion engines

(vi) Transformers

(vii) Condensers

(viii) Evaporators

(ix) Air craft engines

(x) Economizers of steam power plants

(xii) In nature, the ears of Jackrabbits and Fennec Foxes act as fins to release heat from within their bodies.

Fin Performance Parameters

Description of Fin performance in three different ways

(i) Effectiveness of a single fin

(ii) Efficiency of a single fin

(iii)Overall efficiency of an array of fins

FIN EFFECTIVENESS OF A SINGLE FIN

It is applicable for a single fin. It is the ratio of heat transfer with fin to the rate to the heat transfer without fin. Thus it is greater than 1. Practically it should be at least greater than 2.3.

The formula for a fin of infinite length

Єf = q. with fin/ q.without fin= (P h Ac k)1/2(Tb–T)/ h Ac (Tb–T)

= (Pk/h Ac)1/2

Єf = (Pk/h Ac)1/2

Where Ac is the fin cross-sectional area

For a rectangular fin, P = 2B + 2 t and Ac = B t

Where B is width of fin and t is the thickness of fin

Now if the fin is thin, Neglect 2t as compared to 2B

Therefore P/Ac = 2B/B t=2/t

Єf = (2k/ht)1/2 = (1/Bi)0.5

Biot number, Bi for a fin

Less value of Biot number is desirable for more effectiveness.

Biot number will be less if h is small i.e. gas environment all around the fin.

Further, Biot number will be less if thermal conductivity k of fin material is high.

Biot number will be less if the thickness of fin is less i.e. fin is thin.

ASSUMPTIONS USED IN FIN ANALYSIS

One-dimensional steady state heat conduction

Homogeneous and isotropic fin material, k is constant

No internal heat generation

Uniform cross-sectional area

Uniform convective heat transfer coefficient h

Negligible contact thermal resistance

Negligible radiation

Always calculate Rate of heat transfer at the base of the fin

m = (h P/k A_c)^0.5

A_c cross sectional area of the fin if rectangular, A_c =b x t

The fin efficiency is η_f= actual q^._fin/ q^._max with base temperature all along the length

η_f = (PhA_ck)^1/2(T_b–T )/hPL (T_b–T )

= (kA_c/hP)^1/2(1/L)

Finally η_{f =}1/mL

(xii) In nature, the ears of Jackrabbits and Fennec Foxes act as fins to release heat from within their bodies_.

Є_f = q._{with fin}/ q._{without fin}= (P h A_ck)^1/2(T_b–T)/ h A_c (T_b–T)

= (Pk/h A_c)^1/2

Є_f = (Pk/h A_c)^1/2

Where A_cis the fin cross-sectional area

For a rectangular fin, P = 2B + 2 t and A_c = B t

Therefore P/A_c = 2B/B t=2/t

Є_f = (2k/ht)^1/2 = (1/Bi)^0.5

A_c is cross sectional area of the fin = b x t for a rectangular fin

A_sis the surface area of the fin= perimeter x length= (2b+2t) L

d²θ/dx² –m² θ=0

where parameter m = (hP/kA_c)^0.5

q_x = q_x+dx + dq_conv

q_x+dx =q_x+ (dq_x/dx) dx

Firstly q_x= –kA_c (dT/dx)

Secondly q_x+dx = –kA_c (dT/dx) + d(–kA_c (dT/dx) /dx) dx

Simplify q_x+dx = –kA_c (dT/dx) — kA_cd(dT/dx) /dx) dx

q_x+dx = –kA_c (dT/dx) — kA_c d²T/dx²dx

dq_conv = h dA_s (T_x –T_∞)

dA_s= Pdx is the surface area of the differential element

0 = -k A_c (d²T/dx² )dx + h Pdx (T_x –T_∞)

Since A_c is constant, it becomes

d²T/dx² =(hP/kA_c) θ where (T_x –T_∞) =θ & d²T/dx² =d²θ/dx²

d²θ/dx² –m² θ=0

Where fin parameter m = (hP/kA_c)^0.5

θ(x) = C₁e^mx + C₂e^-mx = C₃cosh mx+C₄Sinh mx

Find the constants C₁, C₂, C₃ & C₄ from the boundary conditions.

A_c cross sectional area of the fin if rectangular=b x t

d²θ/dx² –m² θ=0

Where fin parameter m = (hP/kA_c)^0.5

ϴ is the temperature difference θ_x = T_x— T_a

ϴ(x) = C₁e^mx + C₂e^-mx = C₃cosh mx+C₄sinh mx

Find the constants C₁& C₂ (or C₃ and C4)from the boundary conditions.

A_c cross sectional area of the fin if rectangular, A_c =b x t

Boundary conditions (a) at x=0, ϴ=ϴ₀= T_b-T_a

(b) At x = ∞, T =T_a_,and thus ϴ= T_a-T_a =0

(1) Put x=0, C₁ +C₂ = ϴ₀

(ii) Put x =∞, we get C₁ e^∞+ C₂e^-∞ =0 = C₁ e^∞

therefore C₁ =0 and then C₂ = ϴ₀

_{ϴ (x) =ϴ0 e^-mx}

_{ϴ(x)/ ϴ0 = e^-mx}

(T_x –T_a)/( T_b –T_a) = e^–mx